Floating in a Pool

While swimming over break I started thinking about Archimedes Principle and the Physics involved in floating in water and I decided to calculate the buoyant force acting on me while I was floating and submerged underwater. While floating in the water, the buoyant force acting on me is as follows:
ρ(rho)=m/v
Fb=m(fluid)g
When an object is floating the mass of fluid displaced= the mass of the floating object
My mass is about 50 kg so the mass of water displaced= 50kg.
Fb=(50 kg)(9.8 m/s^2)
Fb= 490 N
If I was about 0.5 m underwater, the buoyant force is as follows. For a submerged object the volume of the water displaced= the volume of the object. To calculate my volume, I used two approaches because it is difficult to calculate the volume of a human. First I approximated my body as a cylinder with a radius of about 5 in (12.7 cm or .127 m) and a height of 1.646 m (5 feet 4 inches). The volume of a cylinder is V=πr^2h. Using this formula, my volume is:
V=π(0.127 m)^2(1.646 m)
V=0.0834 m^3= volume of water displaced
My volume can also be calculated using the formula v=m/p
The density of an average human is about 1.0g/cm^3 or 1000 kg/m^3 (interestingly the density of humans is basically equal to the density of water)(http://www.newton.dep.anl.gov/askasci/bio99/bio99503.htm).
Based on this formula:
V= (50 kg)/(1000 kg/m^3)
V=0.05 m^3
I will use the average of these two values, 0.0667 m^3, as the volume of me which equals the volume of water displaced.
Fb=ρvg
Fb= (1000 kg/m^3)(0.0667 m^3)(9.8 m/s^2)
Fb= 653.7 N
The pressure at this depth is:
P=ρgh
P=(1000 kg/m^3)(9.8 m/s^2)(0.5m)
P=4900 N/m^2

I found it interesting to think about the Physics at play while I swim :). Happy Thanksgiving!

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